First determine the atomic number of the ion, and then add or subtract electrons tomatch the charge indicated on the ion. Use the aufbau process to write the completeelectron conﬁguration of the ion.

(a)

Li

+

:Z

=

3, subtract one electron for

+

1charge; 1

s

2

(b)

Ca

2

+

:Z

=

20, subtract two electrons for

+

2 charge; 1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

(c)

Br

−

:Z

=

35, add one electron for charge of

−

1;1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

2

3

d

10

4

p

6

(d)

O

2

−

:Z

=

8, add two electrons for charge of

−

2;1

s

2

2

s

2

2

p

6

Check Your Solution

A check of the sequence of orbitals shows that the aufbau process was followedcorrectly. The number of electrons in each electron conﬁguration matches thedifference of Z

−

(net charge on ion). The answers are correct.

2

.Problem

Draw Lewis structures for the chemical species in question 1.

Solution

In a Lewis structure, the nucleus and the inner core electrons are represented by thesymbol for the element. The electrons in the valence shell are drawn around thesymbol.

(a)(b)(c)(d)

Check Your Solution

Each positive ion shows no electrons. This is expected, since electrons were removedfrom the valence shell. Each negative ion has 8 valence electrons. This is expected forall ions except helium-like ions.

3.

Problem

Draw orbital diagrams and Lewis structures to show how the following pairs of elements can combine. In each case, write the chemical formula for the product.

Solution

In each example, electrons are transferred from the metal to the non-metal so thateach ion attains a noble gas electron conﬁguration.

2

−

••

O

••• •• •

−

••

Br

••• •• •

2+

Ca

+

Li

36

Chapter 4

Structures and Properties of Substances • MHR

CHEMISTRY 12

(a)

Lewis structure:

(b)

Lewis structure:

(c)

Lewis structure:

(d)

Lewis structure:

Check Your Solution

In each case the ions have a noble gas electron con

ﬁ

guration that takes into accountthe charge on the ion. The answers are correct.

→

→

→

→ →→→

1

s

2

s

2

p

N

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

→

3

s

Na

→

→

→

→

→

→

→→

→→

1

s

2

s

2

p

→

3

s

Na

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

→

3

s

Na

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Na

+

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Na

+

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

N

3

−

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Na

+

+

Na

+

Na

+

Na

+

• ••

N

• •• •• •

N

••••

Na

•

Na

•

Na

•

3

−

→

→

→

→

→

→

→→

→→

1

s

2

s

2

p

Cl

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

→

→

→

→

→→

→→

3

s

3

p

→

→

→

→

→

→

→

→

3

s

3

p

→

→

→

→

→

→

→

→

3

s

→

4

s

3

p

→

→

→

→

→

→→

3

s

3

p

K

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Cl

−

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

K

+

+

Cl

− +

• •• •

Cl

• ••

KK

•

→

→

→

→

→

→

→→

→→

1

s

2

s

2

p

→

→

→

→

→

→→

3

s

3

p

Cl

→

→

→

→

→

→

→→

→→

1

s

2

s

2

p

Cl

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

→

→

→

→

→→

→→

3

s

3

p

→

→

→

→

→→

→→

3

s

3

p

→

→

→

→

→

→

→

→

3

s

3

p

→

→

→

→

→

→

→

→

3

s

→

→

4

s

3

p

→

→

→

→

→

→→

3

s

3

p

Ca

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Cl

−

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Cl

−

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

Ca

2

+

+

Ca

−

2

+

• •• •

Cl

• ••• •• •

Cl

• ••

Ca

•• •• •

Cl

••••

−

• •• •

Cl

••••

→

→ →

1

s

2

s

Li

→

→ →

→

→→→

1

s

2

s

3

p

Li

→

→

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

→

→

→

→

→

→

→

→

3

s

3

p

3

s

S

→

→

1

s

Li

+

→

→

1

s

Li

+

→

→

→

→

→

→

→

→

→

→

1

s

2

s

2

p

S

2

−

S

• •••••

Li

+

S

• •• •••••

Li

+

Li

+

••

Li

2

−

37

Chapter 4

Structures and Properties of Substances • MHR

CHEMISTRY 12

4.

Problem

To which main group on the periodic table does X belong?

(a)

MgX

(b)

X

2

SO

4

(c)

X

2

O

3

(d)

XCO

3

Solution

Since these compounds are all ionic, the zero sum rule applies, and the sum of thecharges on the ions must equal zero. Once you know the charge on the ion, it canbe related to the characteristic charge on ions for the main group elements.

(a)

The charge on Mg is

+

2. Therefore, (

+

2)

+

x

= 0, and

x

= −

2. X

2

−

will be ingroup 16.

(b)

The charge onSO

4

is

−

2. Therefore, 2(

x

)

+

(

−

2)

=

0, and

x

=

+

1. X

+

will be ingroup 1.

(c)

The charge on O is

−

2. Therefore, 2(

x

)

+

3(

−

2)

=

0, and

x

= +

3. X

3

+

will be ingroup 3.

(d)

The charge on CO

3

is

−

2. Therefore,

x

+

(

−

2)

=

0, and

x

= +

2. X

2

+

will be ingroup 2.

Check Your Solution

In each case the zero sum rule holds. The charge on each X ion has been determinedcorrectly.

Solutions for Practice Problems

Student Textbook pages 169–170

5.

Problem

List the following compounds in order of decreasing bond energy: H



Br, H



I,H



Cl. Use Appendix E to verify your answer.

Solution

Since all bonds are electrical in nature, a greater difference in charge betweentwo atoms indicates a stronger attraction between the atoms, and therefore, astronger bond. The difference in charge can be estimated using the property of electronegativity. A greater difference in electronegativity between the two atoms inthe bond indicates a greater attraction between the atoms. (A greater electronegativity also indicates a greater ionic character of the bond.) For each bond, measure thedifference in electronegativity, (